I bought it without a rear sway so won't be buying one now. It worked fine without it until recently. Good advice on the shock switch. I'll give that a shot.
Body Roll During Left Turn
- Thread starter steelhd
- Start date
Wouldn't it be asymmetric?
The torsion bar is constrained in all degrees of freedom except to rotate (and torsionally deflect) about its long axis. So the sum of the moments about the bar must be zero. In order to get that with different length arms under load, the short arm must apply more force than the long arm. The difference in the force is proportional to the ratio between the two effective rates. The net moment about the bar axis must be zero, but not necessarily the net force applied to the arms.
In addition, if the springs are equally compressed beyond static, the short arm must maintain a greater angle relative to static than the long arm. As a result, the torsion bar will deflect, and apply a downward force to the short side and a (lesser) upwards force to the long side.
So the net result is that the side with the shorter arm will effectively deflect less than the side with the long bar.
Probably nearly not enough effect to notice, but something interesting to think about.
Only if something is constraining the torsion bar to the frame. If it isn't, then the unequal arm lengths will be equalized through the freely moving torsion bar.
It is constrained in 5 out of 6 degrees of freedom.
Cannot translate (1) up/down or (2) forward/backwards due to the bushings. Cannot translate along it's (3) long axis due to the arms contacting the bushing. Cannot rotate in the (4) roll or (5) yaw directions due to the bushings.
The only degree of freedom the bar has is to (6) rotate in the pitch direction (and deflect along the same axis). There is nothing other than the arms (and a negligible amount of friction) preventing rotation, therefore the moment applied by one arm must equal the moment applied by the other.
Correct. And the force exerted on the axle is going to be the same on each side even if the arms are different lengths.
If the arms are different lengths, then they will provide different force to opposite sides of the axle given the same moment.
In order to generate one foot-pound of moment, a pound can be applied to a lever one foot in length. If you halve the length of the lever, you must double the amount of force to maintain the same moment. So if you have two opposing levers of different lengths, you must apply more force to the shorter lever in order to keep the moment about the fulcrum to be zero.
If you have a fat kid and a skinny kid on a seesaw, the fat kid has to sit closer to the fulcrum than the skinny kid to balance out the seesaw.
If I go outside right now and set the driver's side to the short hole and the passenger side to the long hole and make any adjustments to the link length, the Jeep will not lean from whatever happens to the sway bar. Even if the arms are pointing two different directions.
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Not at rest, no. Because there won't be any moment to create any force. If the force on one arm is zero, the force applied by the other arm is zero.
But deflect the suspension in any way (excluding within a perfect arc that keeps the torsion bar unloaded) and any force the bar does apply will be asymmetric.
I'm going to make a WAG and say that the Antirock is 12" from the torsion bar to the nearest hole and 16" to the farthest hole. Assuming the sum of the moments must be zero (negligible friction):
m1 + m2 = 0, therefore m1=-m2
And moment is length times force times the cosine of the angle between the lever arm and the applied force:
m = L × F × cos(A),
therefore
-L1×F1×cos(A1) = L2×F2×cos(A2).
If we assume that the angles are similar enough to be negligible (right angle approximation), then we get
-L1×F1 = L2×F2
Plugging in L1 is 12 inches and L2 is 16 inches, we get:
F1(12") = -F2(16")
Which we can simplify to:
0.75×F1 = -F2
Therefore, the force applied on the long side will be 3/4 the force applied to the short side. When the force applied on one side is zero, the force applied to the other side is zero. When we apply 1 pound of force to the short side, we get 3/4 pound of force out of the long side.
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Yes. But the fulcrum in in the center. That forces the fat kid to move in. But there is no center fulcrum to the anti rock. That forces the effective force to be equal on both sides.
My apology for the 'fat shaming' I have now lost any hope of one day being President of USA.
It should force the effective moment to be the same on both sides, not the force applied. At the fulcrum, net moment must be zero to prevent rotation.
If we take the seesaw and separate the two sides using a long pipe and two bushings, the requirement for the net moment on the bar to be zero does not change. If we put the arms on the same side of the bar and give one kid negative gravity equivalent to the positive gravity of the other, the fat kid still has to sit closer to the bar.
A more accurate analogy with the seesaw would be if one end was shorter, the kid sitting at that end would have to be fatter to prevent rotation.
Also consider it using limits:
If we give one arm a unit length and force, say 1 foot and 1 pound respectively, and gave the other an infinite length, we can see that as the length of the long arm approaches infinity, applied force from the long arm approaches zero.
If we give the long arm a unit length and force, say 1 foot and 1 pound respectively, then as the shorter arm length approaches zero, the force applied by the short arm approaches infinity.
Imagine you took the Antirock and disconnected the links, but left it in the bushings. On one end you have someone hold the free arm. On the other end you put it in a long pipe and lift it up. The person with the pipe will easily lift the arm on their side, but the person holding the free arm will not be able to restrain the arm from moving.